Optimal. Leaf size=224 \[ \frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}-\frac{i (-b+i a)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{i (b+i a)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.874534, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3567, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}-\frac{i (-b+i a)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{i (b+i a)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3567
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2}{5} \int \frac{-3 a b+\frac{5}{2} \left (a^2-b^2\right ) \tan (c+d x)+2 a b \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{-\frac{3}{4} a \left (5 a^2-b^2\right )-\frac{15}{2} a^2 b \tan (c+d x)-3 a b^2 \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}-\frac{8 \int \frac{\frac{15 a^3 b}{4}-\frac{15}{8} a^2 \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}-\frac{1}{2} \left (i (a-i b)^2\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} \left (i (a+i b)^2\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}-\frac{\left (i (a-i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (i (a+i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}-\frac{\left (i (a-i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left (i (a+i b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac{i (i a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{i (i a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 b \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 a^2-b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 a d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.63091, size = 197, normalized size = 0.88 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (\left (5 a^2-b^2\right ) \tan ^2(c+d x)-a^2-2 a b \tan (c+d x)\right )}{a \tan ^{\frac{5}{2}}(c+d x)}+5 \sqrt [4]{-1} (-a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-5 \sqrt [4]{-1} (a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{5 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.315, size = 1346038, normalized size = 6009.1 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\tan \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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